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Maths NCERT Exemplar Solutions Class 12th Chapter Eight Maths Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Two Functions

Let [t] denote the greatest integer $\leq t$ and {t} denote the fractional part of t. The integral value of for which the left hand limit of the function.

$f (x) = [1 + x] + \frac{α^{2 [x] + {x}} + [x] - 1}{2 [x] + {x}} a t x = 0$ is equal to $α - \frac{4}{3},$ is......................

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Payal Gupta | Contributor-Level 10

2 months ago

$f (x) = [1 + x] + \frac{α^{2 | x |} + {x} + [x] - 1}{2 [x] + {x}}$

$l i m_{x \to 0} - f (x) = α - \frac{4}{3}$

$\Rightarrow l i m_{h \to 0} 1 - 1 + \frac{α^{- h - 1} - 1 - 1}{- h - 1} = α - \frac{4}{3}$

$∴ \frac{α^{- 1} - 2}{- 1} α - \frac{4}{3}$

32 - 10 + 3 = 0

$∴ α = 3 o r 1 / 3$

$? α$ in integer, hence = 3

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Let f : R → R be a function defined by f(x) = 2e²ˣ / (e²ˣ + e). Then f(1/100) + f(2/100) + f(3/100) + .... + f(99/100) is equal to............

Vishal Baghel

f (x) = 2e²? / (e²? +e? ) and f (1-x) = 2e²? / (e²? +e¹? )
∴ f (x) + f (1-x) = 1/2
i.e. f (x) + f (1-x) = 2
∴ f (1/100) + f (2/100) + . + f (99/100)
Σ? f (x/100) + f (1-x/100) + f (1/2)
= 49 x 2 + 1 = 99

Let a be an integer such that lim (x→a) [18 - [1-x]] / [x-3a] exists, where [t] is greatest integer ≤ t. Then a is equal to:

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lim? (x→7) (18- [1-x])/ ( [x-3a])
exist & a∈I.
= lim? (x→7) (17- [-x])/ ( [x]-3a)
exist
RHL = lim? (x→7? ) (17- [-x])/ ( [x]-3a) = 25/ (7-3a) [a ≠ 7/3]
LHL = lim? (x→7? ) (17- [-x])/ ( [x]-3a) = 24/ (6-3a) [a ≠ 2]
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If f(x) = { 1/|x| ; |x| ≥ 1, ax² + b ; |x| < 1 } is differentiable at every point of the domain, then the values of a and b are respectively:

Vishal Baghel

A function f (x) is continuous at x=1, so lim (x→1? ) f (x) = lim (x→1? ) f (x) = f (1).
Assuming a piecewise function like f (x) = { -x, x<1; ax+b, x≥1 } (structure inferred from derivative).
Continuity at x=1: f (1) = 1. a (1)+b = 1 => a+b=1.

The function is differentiable at x=1. The derivative of f (x) at x=1 from the left is -1. The derivative from the right is a.
So, a = -1. (The image has 2a = -1, which would imply a function like -x and ax²+b). Let's assume f' (x) = 2a for x>1.
2a = -1 => a = -1/2.

From a+b=1, b = 1 - a = 1 - (-1/2) = 3/2.
So, a = -1/2 and b = 3/2.

The real function f(x) = arccosec(x) / √(x - [x]), where [x] denotes the greatest integer less than or equal to x, is defined for all x belonging to:

Vishal Baghel

Given the function f (x) = cosec? ¹ (x) / √ {x - [x]} where [x] is the greatest integer function.

The domain of cosec? ¹ (x) is (-∞, -1] U [1, ∞).
For the denominator to be defined, x - [x] ≠ 0, which means {x} ≠ 0 (the fractional part of x is not zero). This implies that x cannot be an integer (x ∉ I).

Combining these conditions, the domain is all non-integer numbers except for those in the interval (-1, 1).

If for x, y $\in$ R, x < 0, y = $l o g_{10} x + l o g_{10} x^{1 / 3} + l o g_{10} x^{1 / 9} + . . . . . u p t o \infty$ terms and $\frac{2 + 4 + 6 + . . . . + 2 y}{3 + 6 + 9 + . . . . + 3 y} = \frac{4}{l o g_{10} x},$ then the ordered pair (x, y) is equal to

Vishal Baghel

$y = l o g_{10} x + l o g_{10} x^{1 / 3} + l o g_{10} x^{1 / 9} + . . . . . . . . u p t o \infty t e r m s$

$= l o g_{10} x (1 + \frac{1}{3} + \frac{1}{9} + . . . . . .)$

y = log₁₀ x × $\frac{1}{1 - \frac{1}{3}} = \frac{3}{2} l o g_{10} x$

Now, $\frac{2 + 4 + 6 + . . . . + 2 y}{3 + 6 + 9 + . . . . + 3 y} = \frac{4}{l o g_{10} x}$

$\Rightarrow \frac{2 \times y \frac{(y + 1)}{2}}{3 y \frac{(y + 1)}{2}} = \frac{4}{l o g_{10} x} s o l o g_{10} x = 6$

$\Rightarrow y = \frac{3}{2} \times 6 = 9$

So, (x, y) = (10⁶, 9)

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Let [t] denote the greatest integer $\leq t$ and {t} denote the fractional part of t. The integral value of for which the left hand limit of the function.

$f (x) = [1 + x] + \frac{α^{2 [x] + {x}} + [x] - 1}{2 [x] + {x}} a t x = 0$ is equal to $α - \frac{4}{3},$ is......................

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Let [t] denote the greatest integer ≤t and {t} denote the fractional part of t. The integral value of for which the left hand limit of the function. f(x)=[1+x]+α2[x]+{x}+[x]−12[x]+{x}at x = 0 is equal to α−43, is......................

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Let [t] denote the greatest integer $\leq t$ and {t} denote the fractional part of t. The integral value of for which the left hand limit of the function.

$f (x) = [1 + x] + \frac{α^{2 [x] + {x}} + [x] - 1}{2 [x] + {x}} a t x = 0$ is equal to $α - \frac{4}{3},$ is......................